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26 Nov 2012, 08:58

6

1

If x is a positive integer, is (x)(x + 2)(x + 4) divisible by 12?

Notice that no matter whether x is even or odd, out of x, x+2 and x+4 one must be a multiple of 3. So, we are basically asked to find whether (x)(x + 2)(x + 4) is divisible by 4. Now, if x=odd, then all three multiples are odd, thus (x)(x + 2)(x + 4) will be odd and not divisible by 4. If x=even, then (x)(x + 2)(x + 4)=even*even*even, thus it'll be divisible by 4.

Therefore, the question boils down to find whether x is even.

(1) x^2 + 2x is a multiple of 3 --> if x=1=odd, then the answer is NO but if x=4, then the answer is YES. Not sufficient.

(2) 3x is a multiple of 2. This statement implies that x=even. Sufficient.

If x is a positive integer, is (x)(x + 2)(x + 4) div by 12?
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14 Nov 2016, 10:37

Bunuel wrote:

Ilomelin wrote:

If x is a positive integer, is (x)(x + 2)(x + 4) divisible by 12?

(1) x2 + 2x is a multiple of 3.

(2) 3x is a multiple of 2.

Merging topics. Please refer to the solution above.

Thank you Bunuel.

Could you please explain me how do we know (other than plugging in numbers) that out of 3 consecutive odd or even integers, one must be divisible by 3? I understand why it works with consecutive integers, but why does it also apply to evenly spaced odds or evens?

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Here's my interpretation, I'd actually love it if someone could tell me if I'm wrong here as I often make little mistakes in the theory on these questions.

Is (x)(x + 2)(x + 4) divisible by 12?

(1) x^2 + 2x is a multiple of 3.(2) 3x is a multiple of 2.

I immediately rephrase the 12 into (2^2)*(3). So we need to be divisible by 2 twice and 3 one. So contained in these three terms we need two multiples of 2 and one multiple of 3.

First observation is we either have consecutive even numbers or consecutive odd numbers. If it's the latter, the answer is definitely a 'no' and if it's the former it will be 'yes'.

Why? Here's a quick aside.

By the way this is not something you have time to do in a question so I would really recommend internalizing this concept as it is the backbone of a lot of problems on divisibility, remainders, and several other question families.

Think about the progression of integers (and this is a good thing to get comfortable with). We need two 2's and one 3 from these three integers. In general, we get a 2 every second integer and we get a 3 every third integer.

Now, if we have three odd numbers above what are we going to get? Three numbers that have ZERO 2's. That's what an odd number is: a number that is not divisible by 2. How many 3's will there be? Well we have three consecutive odd numbers, so three numbers that have a range of 4 and start on an odd number.

For instance:

3, 5, 7 or 13, 15, 17 or 21, 23, 25

We can see from these examples that that is always going to be one number that is a multiple of 3. Will it always be only one multiple of 3? Yes, only one. Why? Well, for the same reason we gave above. Multiples of 3 are going to occur every 3rd integer. Perhaps an easier way to understand this (it helps me) is to visualize it.

Consider three consecutive odd numbers:

5, 7, 9

From zero ascending that looks like this:

0 _ _ _ _ 5 _ 7 _ 9 _ ...

Where are the multiples of 3?

0 _ _ 3 _ 5 6 7 _ 9 _

Algebraically, we can say that for three numbers: x, x + 2, and x +4:

If (x) is a multiple of 3 then we have our multiple of three (and we can say the other two integers are not multiples of 3). If x is NOT a multiple of 3 then either (x + 1) or (x - 1) is, but not both. If (x + 1) is, then so is (x + 4) which is the last of our three numbers. If (x + 1) and (x) are not, then (x - 1) is and so is (x + 2) which is our second integer.

So we can see that for any three consecutive odd integers you are going to hit a multiple of 3 at some point in the three numbers.

Therefore in three consecutive odd integers we will have one multiple of 3 and no multiples of 2.

Now, what if we have three even numbers? Well, obviously there a bunch of 2's. What about 3's, how many will there be? Same logic as above. If x is a multiple of 3 we have one there. If x isn't then either (x + 1) or (x - 1) is. If it's the former then x + 4 is a multiple of 3. If it's (x -1) then (x-2) is a multiple of 3.

Ok so before we go to the statements we've proven (albeit laboriously) that the answer depends entirely on whether x is odd or even.

Statement 1: x^2 + 2x is a multiple of 3.

x*(x + 2) multiple of 3.

We don't know if they are even, this could be 6*8 or 3*5. INSUFFICIENT.

Statement 2: 3x is a multiple of 2.

So if 3x is a multiple of 2, then x must be even because 3*(?) = even then the (?) must be even. SUFFICIENT.

Another way to look at statement 2 is to rephrase the statement into:

3x = 2(i), where i is some integerx = (2/3)*(i)

Therefore for x to remain an integer i must be a multiple of 3. So x is a multiple of 2 and 3. Therefore x is (at a minimum) 6 and we have one 2, one 3 and then our second term in the question will provide the second 2 and we have all we need.

If x is a positive integer, is (x)(x + 2)(x + 4) div by 12?
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27 Aug 2017, 16:37

Bunuel wrote:

If x is a positive integer, is (x)(x + 2)(x + 4) divisible by 12?

Notice that no matter whether x is even or odd, out of x, x+2 and x+4 one must be a multiple of 3. So, we are basically asked to find whether (x)(x + 2)(x + 4) is divisible by 4. Now, if x=odd, then all three multiples are odd, thus (x)(x + 2)(x + 4) will be odd and not divisible by 4. If x=even, then (x)(x + 2)(x + 4)=even*even*even, thus it'll be divisible by 4.

Therefore, the question boils down to find whether x is even.

(1) x^2 + 2x is a multiple of 3 --> if x=1=odd, then the answer is NO but if x=4, then the answer is YES. Not sufficient.

(2) 3x is a multiple of 2. This statement implies that x=even. Sufficient.

Answer: B.

Hope it's clear.

Hi Bunuel - you mentioned the following :

"Notice that no matter whether x is even or odd, out of x, x+2 and x+4 one must be a multiple of 3." - -I AGREE

So, we are basically asked to find whether (x)(x + 2)(x + 4) is divisible by 4. = QUESTION ON THIS STATEMENT

Should it not be 2 out of the 3 is divisible by 4, i.e. we need to find whether x and (x+2) or (x+2) and (x+4) or (X+2) and (x+4) are divisible by 4 ....Given either x or (x+2) or (x+4) is going to be a multiple of 3 .....ONLY THE OTHER TWO have to be multiples of 4 instead ...

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